1. **Problem:** Calculate the net present value (NPV) of a project with a 10% after-tax weighted average cost of capital (WACC), 40% tax rate, and straight-line depreciation. 2. **Formula and rules:** - NPV = \( \sum \frac{CF_t}{(1 + r)^t} - Initial\ Investment \) - After-tax cash flow = (Revenue - Operating Costs - Depreciation) \times (1 - Tax Rate) + Depreciation - Straight-line depreciation = \( \frac{Cost}{Life} \) - WACC is used as discount rate \(r\) 3. **Given:** - Initial cost = 12000 - Life = 3 years - Annual savings = 3600 - Annual revenue increase = 2000 - Tax rate = 0.4 - WACC = 0.1 - Depreciation = \( \frac{12000}{3} = 4000 \) 4. **Calculate after-tax cash flows:** - Earnings before tax = 3600 + 2000 = 5600 - Taxable income = 5600 - 4000 = 1600 - Tax = 1600 \times 0.4 = 640 - Net income = 1600 - 640 = 960 - Add back depreciation (non-cash) = 4000 - After-tax cash flow = 960 + 4000 = 4960 5. **Calculate NPV:** \[ NPV = -12000 + \frac{4960}{(1+0.1)^1} + \frac{4960}{(1+0.1)^2} + \frac{4960}{(1+0.1)^3} \] 6. **Evaluate:** \[ NPV = -12000 + \frac{4960}{1.1} + \frac{4960}{1.21} + \frac{4960}{1.331} \] \[ NPV = -12000 + 4509.09 + 4090.91 + 3724.43 = 1324.43 \] 7. **Answer:** The net present value of the project is approximately \$1324.43. --- 5. **Problem:** Find the first derivative of \( y = x^3 - 12x + 13 \) using the definition of the derivative. 6. **Definition:** \[ \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] 7. **Calculate:** \[ f(x+h) = (x+h)^3 - 12(x+h) + 13 = x^3 + 3x^2h + 3xh^2 + h^3 - 12x - 12h + 13 \] \[ f(x+h) - f(x) = 3x^2h + 3xh^2 + h^3 - 12h \] \[ \frac{f(x+h) - f(x)}{h} = 3x^2 + 3xh + h^2 - 12 \] 8. **Take limit as \( h \to 0 \):** \[ \frac{dy}{dx} = 3x^2 - 12 \] 9. **Answer:** The derivative is \( \frac{dy}{dx} = 3x^2 - 12 \). --- 6. **Problem:** Find relative maxima, minima, and points of inflection of \( y = \frac{x}{\sqrt{1 - x^2}} \). 7. **Domain:** \( -1 < x < 1 \) because of the square root. 8. **First derivative:** \[ y = x(1 - x^2)^{-1/2} \] Using product and chain rules: \[ \frac{dy}{dx} = (1 - x^2)^{-1/2} + x \times \frac{1}{2}(1 - x^2)^{-3/2} \times 2x = (1 - x^2)^{-1/2} + x^2 (1 - x^2)^{-3/2} \] \[ = \frac{1 - x^2 + x^2}{(1 - x^2)^{3/2}} = \frac{1}{(1 - x^2)^{3/2}} \] 9. **Critical points:** \[ \frac{dy}{dx} = 0 \Rightarrow \text{No solution since denominator never zero in domain} \] 10. **Second derivative:** \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( (1 - x^2)^{-3/2} \right) = -\frac{3}{2} (1 - x^2)^{-5/2} (-2x) = 3x (1 - x^2)^{-5/2} \] 11. **Points of inflection:** Set \( \frac{d^2y}{dx^2} = 0 \): \[ 3x (1 - x^2)^{-5/2} = 0 \Rightarrow x = 0 \] 12. **Test concavity:** - For \( x < 0 \), \( \frac{d^2y}{dx^2} < 0 \) (concave down) - For \( x > 0 \), \( \frac{d^2y}{dx^2} > 0 \) (concave up) 13. **Answer:** - No relative maxima or minima in domain. - Point of inflection at \( x = 0 \). --- 7. **Problem:** Sketch the curve \( y = 4 + 3x - x^3 \). 8. **First derivative:** \[ \frac{dy}{dx} = 3 - 3x^2 \] 9. **Critical points:** \[ 3 - 3x^2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \] 10. **Second derivative:** \[ \frac{d^2y}{dx^2} = -6x \] 11. **Classify critical points:** - At \( x = 1 \), \( \frac{d^2y}{dx^2} = -6 < 0 \) max - At \( x = -1 \), \( \frac{d^2y}{dx^2} = 6 > 0 \) min 12. **Answer:** - Relative maximum at \( (1, 4 + 3(1) - 1) = (1, 6) \) - Relative minimum at \( (-1, 4 - 3 - (-1)) = (-1, 2) \) --- 8. **Problem:** MMC furniture revenue function \( R = 24x - 3x^2 \). (a) Max revenue: - Derivative \( R' = 24 - 6x \) - Set \( R' = 0 \Rightarrow 24 - 6x = 0 \Rightarrow x = 4 \) - Max revenue \( R(4) = 24(4) - 3(16) = 96 - 48 = 48 \) (b) Average revenue: \[ AR = \frac{R}{x} = \frac{24x - 3x^2}{x} = 24 - 3x \] (c) Marginal revenue: \[ MR = R' = 24 - 6x \] (d) Plotting all three on one graph (not shown here). --- 9. **Problem:** Demand \( y = 30 - 2x^2 \), Supply \( y = 3 + x^2 \), find max revenue from tax \( t \) per unit and tax rate. - Equilibrium without tax: \( 30 - 2x^2 = 3 + x^2 \Rightarrow 27 = 3x^2 \Rightarrow x = 3 \) - Tax shifts supply: new supply \( y = 3 + x^2 + t \) - New equilibrium: \[ 30 - 2x^2 = 3 + x^2 + t \Rightarrow 27 - t = 3x^2 \Rightarrow x = \sqrt{\frac{27 - t}{3}} = \sqrt{9 - \frac{t}{3}} \] - Revenue from tax \( R = t \times x = t \sqrt{9 - \frac{t}{3}} \) - Maximize \( R(t) \) by derivative: \[ \frac{dR}{dt} = \sqrt{9 - \frac{t}{3}} + t \times \frac{-1}{6 \sqrt{9 - \frac{t}{3}}} = 0 \] - Solve for \( t \) gives max tax revenue. --- 10. **Problem:** Marginal cost \( MC = 10 + 24x - 3x^2 \), total cost at 1 unit is 25. - Total cost \( C(x) = \int MC dx = 10x + 12x^2 - x^3 + C_0 \) - Use \( C(1) = 25 \): \[ 10(1) + 12(1)^2 - 1^3 + C_0 = 25 \Rightarrow 10 + 12 - 1 + C_0 = 25 \Rightarrow C_0 = 4 \] - Total cost function: \[ C(x) = 10x + 12x^2 - x^3 + 4 \] - Average cost: \[ AC = \frac{C(x)}{x} = 10 + 12x - x^2 + \frac{4}{x} \] ---